what is the connected component of $\Bbb{C}^*$ what is the connected component of $\Bbb{C}^*$ ? is it $\Bbb{C}^*$ ? 
where $\Bbb{C}^*=\Bbb{C}-\{0\}$ 
Thanks in advance
 A: Observe that $\mathbb{C} \setminus \{0\}$ is homeomorphic to $\mathbb{R}^2 \setminus \{0\}$. 
Now we can easily show that $\mathbb{R}^2 \setminus \{0\}$ is path connected. Pick $x, y \in \mathbb{R}^2 \setminus \{0\}$ such that $x \neq y$ and let $z \in \mathbb{R}^2 \setminus \{0\}$ such that $z \neq x$ and $z \neq y$.
Then define $f : [0, 1] \to \mathbb{R}^2 \setminus \{0\}$ by $f(a) = (1-a)x + tz$, then $f$ represents the straight line path between $x$ and $z$. Similarly define $g : [0, 1] \to \mathbb{R}^2 \setminus \{0\}$ by $g(a) = (1-a)z + ty$, then $g$ represents the straight line path between $z$ and $y$. Now define $h : [0, 1] \to \mathbb{R}^2 \setminus \{0\}$ by $$h(a) =
\begin{cases}
f(2a) \ \ \ \ \text{if} \ \ a \in [0, \frac{1}{2}]\\
g(2a) \ \ \ \ \text{if} \ \ a \in [\frac{1}{2}, 1]\\
\end{cases}$$
Then since $g(\frac{1}{2}) = f(\frac{1}{2}) = z$ we have $h$ to be continuous by the gluing lemma  and $h(0) = f(0) = x$ and $h(1) = g(1) = y$ and thus $h$ is a path from $x$ to $y$ in $\mathbb{R}^2 \setminus \{0\}$ thus $\mathbb{R}^2 \setminus \{0\}$ is path-connected and thus connected.
So $\mathbb{C} \setminus \{0\}$ must be connected since homeomorphisms preserve connectedness. And thus $\mathbb{C} \setminus \{0\}$  is the only connected component of $\mathbb{C} \setminus \{0\}$.
